Hong Kong Preliminary Selection Contest

Since $a$ is a root of $g(x)=0$, we have $a^4-a^3-a^2-1=0$. Using this relation, we get $$f(a)=a^6-a^5-a^3-a^2-a=(a^2+1)(a^4-a^3-a^2-1)+a^2-a+1=a^2-a+1.$$ Of course, the same also applies to $b,c$ and $d$. Therefore, we have $$f(a)+f(b)+f(c)+f(d)=(a^2+b^2+c^2+d^2)-(a+b+c+d)+4.$$ Since $a,b,c,d$ are the roots of the equation $x^4-x^3-x^2-1=0$, we have $a+b+c+d=1$ and $ab+ac+ad+bc+bd+cd=-1$. It follows that $$a^2+b^2+c^2+d^2=(a+b+c+d{)}^2-2(ab+ac+ad+bc+bd+cd)=3,$$ and so $f(a)+f(b)+f(c)+f(d)=3-1+4=6$.